5
$\begingroup$

Is arc connected-ness well-behaved with respect to products?

That is -

$\prod X_\alpha$ is arc connected iff $X_\alpha$ is arc connected $\forall \alpha$

In this question on MathStackexchange, an answer is provided only for the reverse implication, that is -

If $X_\alpha$ is arc connected $\forall \alpha$, then $\prod X_\alpha$ is arc connected

However, I wasn't able to get an answer for the forward implication, nor have I been able to find it in any book or from Googling. So, is the forward implication true, or is there a counterexample for the same?

$\endgroup$
2
  • $\begingroup$ They are equivalent when $X_\alpha$ is Hausdorff because in this context path-connected implies arc-connected. I will now assume that you are asking about T$_1$ spaces. $\endgroup$ Commented Aug 8, 2020 at 19:00
  • $\begingroup$ In fact, I'm asking for general $X_\alpha$, which need not even be $T_0$. $\endgroup$
    – Ishan Deo
    Commented Aug 8, 2020 at 19:01

1 Answer 1

7
$\begingroup$

This is false when the spaces are not Hausdorff. Let $X$ be the line with two origins $\{O_1,O_2\}$ and $Y$ be the usual line. Then $X\times Y$ is arc connected because you can pick an arc that starts at $(O_1,y_1)$ travels outside $\{O_1,O_2\}\times Y$ and then comes back to $(O_2,y_2)$, but $X$ itself is not arc connected.

$\endgroup$
4
  • $\begingroup$ Sorry it's early in the morning and I might be missing something obvious, but writing $X=\mathbb{R}\sqcup\mathbb{R}/\sim$ why can't I take the path $[0,\frac{1}{2}]\to[0,\frac{1}{2}]$ on the first copy of $\mathbb{R}$ and $[\frac{1}{2},1]\to[\frac{1}{2},0]$ on the second copy to link the two origins? $\endgroup$ Commented Aug 9, 2020 at 6:17
  • $\begingroup$ @DanielRobert-Nicoud that shows that it is path connected, which is different from arc connected. You would need your map from $[0,1]\to X$ to be injective. $\endgroup$ Commented Aug 9, 2020 at 6:21
  • $\begingroup$ Ah I see! Thanks, I had never come across this notion tbh. $\endgroup$ Commented Aug 9, 2020 at 6:24
  • $\begingroup$ Also since the projection of a path is still a path, the implication of the OP is true for path connected.. $\endgroup$ Commented Aug 9, 2020 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.